Show that : $\sum_{k=0}^{n}{(-1)^k{n\choose k}\over (k+1)^2}={1+\sqrt{1+4(n+1)^2H_nH_{n+1}}\over 2(n+1)^2}$, for $n\ge0$

Question

Show that

$$\sum_{k=0}^{n}{(-1)^k{n\choose k}\over (k+1)^2}={1+\sqrt{1+4(n+1)^2H_nH_{n+1}}\over 2(n+1)^2}$$

$H_n$; is the n-th harmonic number

My try:

We know that

$$\sum_{k=0}^{\infty}{(-1)^k\over (k+1)^2}=\int_{0}^{1}{-\ln{x}\over 1+x}dx$$

But

$$\sum_{k=0}^{n}{(-1)^k{n\choose k}\over (k+1)^2}=\int_{0}^{a}???$$

From wikipedia I found

$$H_n=\int_{0}^{1}{1-x^n\over 1-x}dx=\sum_{k=1}^{n}(-1)^{k-1}{{n\choose k}\over k}$$ may be of useful

Onward I am clueless as what to do next. Please help!

Answer 1

Start by the binomial expansion

$$\sum^n_{k=0} {n \choose k} x^k = (1+x)^n$$

By integration

$$\sum^n_{k=0} {n \choose k} \frac{x^{k+1}}{k+1} =\frac{(1+x)^{n+1}-1}{n+1}$$

Integrating again

$$\sum^n_{k=0} {n \choose k} \frac{(-1)^{k+2}}{(k+1)^2} =\frac{1}{n+1}\int^{-1}_0\frac{(1+x)^{n+1}-1}{x}\,dx = \frac{1}{n+1}\int^{1}_0\frac{(1-x)^{n+1}-1}{x}\,dx$$

By substitution

$$\sum^n_{k=0} {n \choose k} \frac{(-1)^{k}}{(k+1)^2} =\frac{1}{n+1}\int^{1}_0\frac{x^{n+1}-1}{x-1}\,dx = \frac{H_{n+1}}{n+1} $$

Note that

\begin{align} \frac{1+\sqrt{1+4(n+1)^2H_nH_{n+1}}}{n+1} &= \frac{1+\sqrt{1+4(n+1)^2H_n^2+4(n+1)H_n}}{n+1}\\&=\frac{ 1+\sqrt{(1+2(n+1)H_n)^2}}{n+1} \\&=\frac{2+2(n+1)H_n}{2(n+1)^2} \\&= \frac{H_{n+1}}{n+1} \end{align}

Answer 2

Since $\frac{1}{(1+k)^2}=\int_{0}^{1} x^k(-\log x)\,dx $ we have:

$$\begin{eqnarray*} \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{(k+1)^2} &=& \int_{0}^{1}(-\log x)\sum_{k=0}^{n}\binom{n}{k}(-x)^k\,dx\\ &=&\int_{0}^{1}(-\log x)(1-x)^n\,dx\\&=&-\int_{0}^{1}x^n\log(1-x)\,dx\\&=&-\left.\frac{d}{d\alpha}\int_{0}^{1}x^n(1-x)^{\alpha}\,dx\,\right|_{\alpha=0^+}\\&=&-\left.\frac{d}{d\alpha}\frac{\Gamma(n+1)\,\Gamma(\alpha+1)}{\Gamma(n+\alpha+1)}\,\right|_{\alpha=0^+}\\&=&\frac{H_{n+1}}{n+1} \end{eqnarray*}$$ and since $H_{n+1}=H_n+\frac{1}{n+1}$, $$ 1+4(n+1)^2 H_n H_{n+1} = 1+4(n+1) H_n + 4(n+1)^2 H_n^2 = (1+2(n+1)H_n)^2.$$